Partial Fractions — Distinct Linear Factors

Suppose $Q(x) = (x - a_1)(x - a_2)\cdots(x - a_n)$ with all $a_i$ distinct, and $\deg P < n$. Then there exist unique constants $A_1, A_2, \dots, A_n$ such that

Cover-up method. To find $A_k$: multiply both sides by $(x - a_k)$, then substitute $x = a_k$. On the left, $(x - a_k)$ cancels, leaving $P(a_k)/\prod_{i \neq k}(a_k - a_i)$. On the right, every term except $A_k$ has a factor of $(x - a_k)$ in its denominator and survives — but they vanish when $x = a_k$ is substituted into the cleared form. So

Working the hook: $\dfrac{1}{(x-1)(x+1)} = \dfrac{A}{x-1} + \dfrac{B}{x+1}$. Cover $(x-1)$ and set $x = 1$: $A = \dfrac{1}{1 + 1} = \dfrac{1}{2}$. Cover $(x+1)$ and set $x = -1$: $B = \dfrac{1}{-1 - 1} = -\dfrac{1}{2}$.

Template for distinct linear factors: one $A_i/(x - a_i)$ per root · Cover-up: $A_k = P(a_k) / \prod_{i \neq k}(a_k - a_i)$ · $\int 1/(x-a)\,dx = \ln|x-a| + C$ — keep the absolute value · Requirement: $\deg P < \deg Q$ (divide first if not)

Pause — copy the distinct-linear-factor template, the cover-up formula $A_k = P(a_k)/\prod_{i\neq k}(a_k-a_i)$, and the integral $\int 1/(x-a)\,dx = \ln|x-a|+C$ into your book.

We just saw the distinct-linear-factor template (one $A_i/(x-a_i)$ per root) and the cover-up rule $A_k = P(a_k)/\prod_{i\neq k}(a_k-a_i)$, integrating to a sum of $\ln|x-a_i|$ terms. That raises a question: what if cover-up is unavailable — say, for the constant in a sum of two terms? This card answers it → clear denominators, expand, and equate coefficients of each power of $x$ to get a linear system.

Cover-up is fastest for distinct linear factors, but you should also master the equating coefficients method — it generalises to every case (repeated factors, quadratics) you'll meet in L06.

Procedure for $\dfrac{P(x)}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}$:

• Multiply both sides by $(x-a)(x-b)$ to clear denominators: $P(x) = A(x-b) + B(x-a)$.
• Expand the right side and collect powers of $x$.
• Match the coefficient of each power of $x$ on the left to that on the right — gives a linear system for $A$, $B$.
• Or: substitute strategic $x$-values (especially the roots) to isolate one constant at a time — this is the cover-up shortcut in disguise.

Equating coefficients: clear denominators, match powers of $x$ · Or substitute roots to isolate constants (cover-up reinterpreted) · If improper ($\deg P \geq \deg Q$), divide first · Check by recombining your decomposition over a common denominator

Pause — copy the equating-coefficients method (clear denominators, match powers of $x$), the improper-fraction check ($\deg P \geq \deg Q$ → divide first), and the verification by recombining into your book.